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Ch. 3 Solutions

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Basic

Medium

Advanced

Basic

B-1. Euler-Lagrange Equation from the Lagrangian Density (Klein-Gordon Field)

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(a) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial \phi}\)

In \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2\), the only term containing \(\phi\) itself (without derivatives) is \(-\frac{1}{2}m^2\phi^2\).

\[ \boxed{\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi} \]

(b) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\)

Rewriting the kinetic term as \(\frac{1}{2}\eta^{\alpha\beta}\partial_\alpha\phi\,\partial_\beta\phi\) and differentiating with respect to \(\partial_\mu\phi\):

\[ \frac{\partial}{\partial(\partial_\mu\phi)}\left[\frac{1}{2}\eta^{\alpha\beta}\partial_\alpha\phi\,\partial_\beta\phi\right] = \frac{1}{2}\eta^{\alpha\beta}(\delta^\mu_\alpha\,\partial_\beta\phi + \partial_\alpha\phi\,\delta^\mu_\beta) = \frac{1}{2}(\eta^{\mu\beta}\partial_\beta\phi + \eta^{\alpha\mu}\partial_\alpha\phi) = \partial^\mu\phi \]
\[ \boxed{\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} = \partial^\mu\phi} \]

(c) Derivation of the Klein-Gordon equation

Substituting into the Euler-Lagrange equation \(\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial\mathcal{L}}{\partial\phi} = 0\):

\[ \partial_\mu(\partial^\mu\phi) - (-m^2\phi) = 0 \]
\[ \boxed{\partial_\mu\partial^\mu\phi + m^2\phi = 0} \]

This is the Klein-Gordon equation.

Verification: Substituting a plane wave \(\phi \propto e^{-ik\cdot x}\) gives \(-k_\mu k^\mu + m^2 = 0\), i.e., \(E^2 = |\mathbf{p}|^2 + m^2\), which is consistent with the relativistic dispersion relation.

B-2. Expanding the Indices of the Kinetic Term

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Using \(\partial^\mu\phi = \eta^{\mu\nu}\partial_\nu\phi\), we verify each component:

\[ \partial^0\phi = \eta^{00}\partial_0\phi = (+1)\partial_0\phi = \dot{\phi} \]
\[ \partial^i\phi = \eta^{ii}\partial_i\phi = (-1)\partial_i\phi = -\partial_i\phi \quad (i = 1, 2, 3) \]

Summing over \(\mu\):

\[ \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi = \frac{1}{2}\left[\partial_0\phi\,\partial^0\phi + \sum_{i=1}^{3}\partial_i\phi\,\partial^i\phi\right] \]
\[ = \frac{1}{2}\left[\dot{\phi}\cdot\dot{\phi} + \sum_{i=1}^{3}\partial_i\phi\cdot(-\partial_i\phi)\right] \]
\[ \boxed{\frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2} \]

Verification: From the metric signature \((+,-,-,-)\), we can confirm that the temporal component carries a positive sign and the spatial components carry negative signs.

B-3. Dispersion Relation for a Massless Field

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The equation of motion obtained from the Lagrangian density \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi\) with \(m = 0\) is:

\[ \partial_\mu\partial^\mu\phi = 0 \quad (\text{wave equation}) \]

Computing each derivative of the plane wave solution \(\phi(x) = A\,e^{-iEt + i\mathbf{p}\cdot\mathbf{x}}\):

\[ \partial_0\phi = -iE\,\phi, \qquad \partial_0^2\phi = (-iE)^2\phi = -E^2\phi \]
\[ \partial_i\phi = ip_i\,\phi, \qquad \nabla^2\phi = (ip_1)^2\phi + (ip_2)^2\phi + (ip_3)^2\phi = -|\mathbf{p}|^2\phi \]

Computing the d'Alembertian:

\[ \partial_\mu\partial^\mu\phi = \partial_0^2\phi - \nabla^2\phi = -E^2\phi - (-|\mathbf{p}|^2\phi) = (-E^2 + |\mathbf{p}|^2)\phi = 0 \]

Since \(\phi \neq 0\):

\[ \boxed{E^2 = |\mathbf{p}|^2} \]

Verification: This corresponds to the dispersion relation \(E = |\mathbf{p}|\) for massless particles (such as photons), which is physically correct.

B-4. Lagrangian with Interaction Term

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\[ \mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4 \]

Step 1: Partial derivative with respect to \(\phi\)

\[ \frac{\partial\mathcal{L}}{\partial\phi} = -m^2\phi - \frac{\lambda}{4!}\cdot 4\phi^3 = -m^2\phi - \frac{\lambda}{3!}\phi^3 \]

Step 2: Partial derivative with respect to \(\partial_\mu\phi\)

The \(\phi^4\) term does not contain \(\partial_\mu\phi\), so the result is the same as D1(b):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi \]

Step 3: Substitution into the Euler-Lagrange equation

\[ \partial_\mu(\partial^\mu\phi) - \left(-m^2\phi - \frac{\lambda}{3!}\phi^3\right) = 0 \]
\[ \boxed{(\partial_\mu\partial^\mu + m^2)\phi + \frac{\lambda}{3!}\phi^3 = 0} \]

Verification: Setting \(\lambda = 0\) recovers the free Klein-Gordon equation. Also, differentiating \(\frac{\lambda}{4!}\phi^4\) with respect to \(\phi\) gives \(\frac{4\lambda}{4!}\phi^3 = \frac{\lambda}{3!}\phi^3\), confirming that the factorial manipulation is correct.

B-5. Component Expression of the d'Alembertian

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Expanding \(\Box \equiv \partial_\mu\partial^\mu = \eta^{\mu\nu}\partial_\mu\partial_\nu\):

\[ \Box = \eta^{00}\partial_0\partial_0 + \eta^{11}\partial_1\partial_1 + \eta^{22}\partial_2\partial_2 + \eta^{33}\partial_3\partial_3 \]
\[ = (+1)\frac{\partial^2}{\partial t^2} + (-1)\frac{\partial^2}{\partial x^2} + (-1)\frac{\partial^2}{\partial y^2} + (-1)\frac{\partial^2}{\partial z^2} \]
\[ \boxed{\Box = \frac{\partial^2}{\partial t^2} - \nabla^2 = \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2}} \]

Writing the Klein-Gordon equation \((\Box + m^2)\phi = 0\) in components:

\[ \boxed{\frac{\partial^2\phi}{\partial t^2} - \frac{\partial^2\phi}{\partial x^2} - \frac{\partial^2\phi}{\partial y^2} - \frac{\partial^2\phi}{\partial z^2} + m^2\phi = 0} \]

Verification: Setting \(m = 0\) reduces this to the ordinary wave equation \(\frac{\partial^2\phi}{\partial t^2} = \nabla^2\phi\).

B-6. Component Expansion of the Continuity Equation

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(a) Continuity Equation

Expanding \(\partial_\mu j^\mu = 0\):

\[ \partial_\mu j^\mu = \partial_0 j^0 + \partial_1 j^1 + \partial_2 j^2 + \partial_3 j^3 = \frac{\partial j^0}{\partial t} + \nabla\cdot\mathbf{j} = 0 \]
\[ \boxed{\frac{\partial j^0}{\partial t} + \nabla\cdot\mathbf{j} = 0} \]

This has the same form as the continuity equation for charge density \(\rho = j^0\) and current density \(\mathbf{j}\).

(b) Time Independence of the Conserved Charge

\[ \frac{dQ}{dt} = \frac{d}{dt}\int d^3x\,j^0 = \int d^3x\,\frac{\partial j^0}{\partial t} \]

Substituting the result from (a), \(\frac{\partial j^0}{\partial t} = -\nabla\cdot\mathbf{j}\):

\[ \frac{dQ}{dt} = -\int d^3x\,\nabla\cdot\mathbf{j} \]

Applying Gauss's theorem:

\[ \frac{dQ}{dt} = -\oint_{\partial V}\mathbf{j}\cdot d\mathbf{S} \]

Assuming that \(\mathbf{j}\) falls off sufficiently rapidly at spatial infinity (\(|\mathbf{j}| \to 0\) as \(|\mathbf{x}| \to \infty\)), the surface integral vanishes when the integration region is taken over all space:

\[ \boxed{\frac{dQ}{dt} = 0} \]

Therefore, \(Q\) is a conserved quantity that does not depend on time.

Consistency check: As a dimensional analysis check, if \(j^0\) has the dimensions of charge density, then \(Q = \int d^3x\,j^0\) has the dimensions of charge, which is consistent.

B-7. Partial Derivatives of the Complex Scalar Field

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\[ \mathcal{L} = \partial_\mu\phi^*\,\partial^\mu\phi - m^2\phi^*\phi = \eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi - m^2\phi^*\phi \]

(a) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial \phi^*}\)

The only term containing \(\phi^*\) itself (without derivatives) is \(-m^2\phi^*\phi\). We treat \(\phi\) as a constant:

\[ \boxed{\frac{\partial\mathcal{L}}{\partial\phi^*} = -m^2\phi} \]

(b) Calculation of \(\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi^*)}\)

We differentiate \(\mathcal{L} = \eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi - m^2\phi^*\phi\) with respect to \(\partial_\mu\phi^*\). Since \(\partial_\beta\phi\) does not depend on \(\partial_\mu\phi^*\), it can be treated as a constant:

\[ \frac{\partial}{\partial(\partial_\mu\phi^*)}\left[\eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi\right] = \eta^{\alpha\beta}\delta^\mu_\alpha\,\partial_\beta\phi = \eta^{\mu\beta}\partial_\beta\phi = \partial^\mu\phi \]
\[ \boxed{\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)} = \partial^\mu\phi} \]

(c) Euler-Lagrange equation for \(\phi^*\)

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)}\right) - \frac{\partial\mathcal{L}}{\partial\phi^*} = 0 \]
\[ \partial_\mu(\partial^\mu\phi) - (-m^2\phi) = 0 \]
\[ \boxed{(\partial_\mu\partial^\mu + m^2)\phi = 0} \]

The Klein-Gordon equation for \(\phi\) is obtained.

Verification: Similarly, deriving the Euler-Lagrange equation for \(\phi\) yields \((\partial_\mu\partial^\mu + m^2)\phi^* = 0\), confirming that \(\phi\) and \(\phi^*\) each independently satisfy the Klein-Gordon equation.

B-8. Construction of Noether Current (Application of the Formula)

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(a) Calculation of \(\delta\mathcal{L}\)

Under a constant infinitesimal shift \(\delta\phi = \epsilon\), we have \(\partial_\mu(\delta\phi) = 0\), so \(\delta(\partial_\mu\phi) = 0\).

\[ \delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta(\partial_\mu\phi) = (-m^2\phi)\cdot\epsilon + \partial^\mu\phi\cdot 0 = -m^2\phi\,\epsilon \]

Since \(\delta\mathcal{L} \neq 0\) when \(m \neq 0\) and \(\phi \neq 0\):

\[ \boxed{\text{For } m \neq 0\text{, the constant shift is not a symmetry}} \]

(b) The case \(m = 0\)

When \(m = 0\):

\[ \delta\mathcal{L} = -0\cdot\phi\,\epsilon = 0 \]

Therefore the constant shift is a symmetry. The corresponding conserved current is:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi = \partial^\mu\phi\cdot\epsilon \]

Factoring out \(\epsilon\) (since \(\epsilon\) is an arbitrary infinitesimal constant, we extract the coefficient):

\[ \boxed{j^\mu = \partial^\mu\phi} \]

Verification: The equation of motion for \(m = 0\) is \(\partial_\mu\partial^\mu\phi = 0\), so \(\partial_\mu j^\mu = \partial_\mu\partial^\mu\phi = 0\) is indeed satisfied. For \(m \neq 0\), we have \(\partial_\mu j^\mu = \partial_\mu\partial^\mu\phi = -m^2\phi \neq 0\), which is consistent with the current not being conserved.

Medium

M-1. Derivation of Conjugate Momentum and Hamiltonian Density

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(a) Conjugate Momentum Density

\(\mathcal{L} = \frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2\) (using the result from D2)

\[ \pi(x) \equiv \frac{\partial\mathcal{L}}{\partial\dot{\phi}} = \frac{\partial}{\partial\dot{\phi}}\left[\frac{1}{2}\dot{\phi}^2\right] = \dot{\phi} \]
\[ \boxed{\pi(x) = \dot{\phi}(x)} \]

(b) Hamiltonian Density

We compute the Legendre transformation \(\mathcal{H} = \pi\dot{\phi} - \mathcal{L}\). Using \(\dot{\phi} = \pi\):

\[ \mathcal{H} = \pi\cdot\pi - \left[\frac{1}{2}\pi^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2\right] \]
\[ = \pi^2 - \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]
\[ \boxed{\mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2} \]

(c) Physical Interpretation

By analogy with the particle mechanics Hamiltonian \(H = T + V\) (kinetic energy + potential energy):

Term Physical meaning Correspondence to particle mechanics
\(\frac{1}{2}\pi^2 = \frac{1}{2}\dot{\phi}^2\) Energy density associated with the time variation of the field Kinetic energy \(\frac{1}{2}m\dot{q}^2\)
\(\frac{1}{2}(\nabla\phi)^2\) Energy density associated with spatial gradients of the field (the cost of neighboring points having different field values) Spring potential energy in a system of masses connected by springs
\(\frac{1}{2}m^2\phi^2\) Energy density associated with the field value itself (mass term) Harmonic oscillator potential \(\frac{1}{2}k q^2\)

An important point is that all three terms in \(\mathcal{H}\) are positive definite, so the energy density satisfies \(\mathcal{H} \geq 0\). This guarantees the stability of the theory.

Consistency check: We verify the dimensions of \(\mathcal{H}\). In natural units, \([\phi] = \text{mass}^1\) and \([\partial_\mu] = \text{mass}^1\), so \([\mathcal{H}] = \text{mass}^4\), which matches the dimensions of energy density (= energy/volume = mass\(^4\)).

M-2. Internal Symmetry and Noether Current of the Complex Scalar Field

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(a) Verification that \(\delta\mathcal{L} = 0\)

Under the infinitesimal transformation \(\delta\phi = i\alpha\phi\), \(\delta\phi^* = -i\alpha\phi^*\):

\[ \delta(\partial_\mu\phi) = i\alpha\,\partial_\mu\phi, \qquad \delta(\partial_\mu\phi^*) = -i\alpha\,\partial_\mu\phi^* \]

Computing the variation of \(\mathcal{L} = \partial_\mu\phi^*\,\partial^\mu\phi - m^2\phi^*\phi\):

\[ \delta\mathcal{L} = \delta(\partial_\mu\phi^*)\,\partial^\mu\phi + \partial_\mu\phi^*\,\delta(\partial^\mu\phi) - m^2[\delta(\phi^*)\,\phi + \phi^*\,\delta\phi] \]

Substituting each term:

\[ = (-i\alpha\,\partial_\mu\phi^*)\,\partial^\mu\phi + \partial_\mu\phi^*\,(i\alpha\,\partial^\mu\phi) - m^2[(-i\alpha\phi^*)\phi + \phi^*(i\alpha\phi)] \]
\[ = -i\alpha\,\partial_\mu\phi^*\,\partial^\mu\phi + i\alpha\,\partial_\mu\phi^*\,\partial^\mu\phi - m^2[-i\alpha\phi^*\phi + i\alpha\phi^*\phi] \]
\[ = 0 + 0 = 0 \]
\[ \boxed{\delta\mathcal{L} = 0} \]

(b) Derivation of the Noether Current

For a complex scalar field, the Noether current is:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)}\,\delta\phi^* \]

Computing each partial derivative. From \(\mathcal{L} = \eta^{\alpha\beta}\partial_\alpha\phi^*\,\partial_\beta\phi - m^2\phi^*\phi\):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi^* \]
\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^*)} = \partial^\mu\phi \]

Substituting:

\[ j^\mu = \partial^\mu\phi^*\cdot(i\alpha\phi) + \partial^\mu\phi\cdot(-i\alpha\phi^*) \]
\[ = i\alpha\left[\phi\,\partial^\mu\phi^* - \phi^*\,\partial^\mu\phi\right] \]

Extracting the coefficient of \(\alpha\) to define the conserved current:

\[ \boxed{j^\mu = i\left(\phi\,\partial^\mu\phi^* - \phi^*\,\partial^\mu\phi\right)} \]

(c) Conserved Charge

\[ Q = \int d^3x\,j^0 = i\int d^3x\left(\phi\,\partial^0\phi^* - \phi^*\,\partial^0\phi\right) = i\int d^3x\left(\phi\,\dot{\phi}^* - \phi^*\,\dot{\phi}\right) \]
\[ \boxed{Q = i\int d^3x\left(\phi\,\dot{\phi}^* - \phi^*\,\dot{\phi}\right)} \]

Physical meaning: After quantization, the complex scalar field describes both particles and antiparticles. When the field \(\phi\) is Fourier expanded, the positive-frequency part corresponds to particle annihilation operators, while the negative-frequency part corresponds to antiparticle creation operators. After quantization, the conserved charge \(Q\) becomes an operator proportional to "particle number \(-\) antiparticle number" (\(N_{\text{particle}} - N_{\text{antiparticle}}\)). This is a conserved quantity corresponding to \(U(1)\) symmetry, and when coupled to electromagnetic interactions, it corresponds to charge conservation.

Consistency check: If \(\phi\) is a real scalar field (\(\phi = \phi^*\)), then \(j^\mu = i(\phi\,\partial^\mu\phi - \phi\,\partial^\mu\phi) = 0\), which is consistent with the fact that a real scalar field carries no \(U(1)\) charge.

M-3. Spacetime Translational Invariance and the Energy-Momentum Tensor

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(a) Proof that \(\delta\mathcal{L}\) is a total derivative

Since \(\mathcal{L}(\phi, \partial_\mu\phi)\) does not depend explicitly on \(x^\mu\), the spacetime dependence of \(\mathcal{L}\) arises only through the field \(\phi(x)\). Under a spacetime translation \(x^\mu \to x^\mu + a^\mu\), the field transforms as \(\delta\phi = -a^\nu\partial_\nu\phi\).

We compute the variation of \(\mathcal{L}\) directly. Viewing \(\mathcal{L}\) as a composite function of \(x\):

\[ \delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta(\partial_\mu\phi) \]

On the other hand, since \(\mathcal{L}\) itself is a scalar field, under the same translation:

\[ \delta\mathcal{L} = -a^\nu\partial_\nu\mathcal{L} \]

We rewrite this in the form of a total derivative. Since \(a^\nu\) is constant:

\[ \delta\mathcal{L} = -a^\nu\partial_\nu\mathcal{L} = -\partial_\mu(a^\nu\delta^\mu{}_\nu\mathcal{L}) = \partial_\mu(-a^\nu\delta^\mu{}_\nu\mathcal{L}) \]

Here we used \(\partial_\mu\delta^\mu{}_\nu = 0\) (constant) and \(\partial_\mu a^\nu = 0\) (constant).

\[ \boxed{\delta\mathcal{L} = \partial_\mu(-a^\nu\delta^\mu{}_\nu\mathcal{L})} \]

This is of the form \(\delta\mathcal{L} = \partial_\mu K^\mu\), where \(K^\mu = -a^\nu\delta^\mu{}_\nu\mathcal{L}\).

(b) Derivation of the energy-momentum tensor

According to the generalized Noether's theorem (for the case \(\delta\mathcal{L} = \partial_\mu K^\mu\)), the conserved current is:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu \]

Substituting \(\delta\phi = -a^\nu\partial_\nu\phi\) and \(K^\mu = -a^\nu\delta^\mu{}_\nu\mathcal{L}\):

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,(-a^\nu\partial_\nu\phi) - (-a^\nu\delta^\mu{}_\nu\mathcal{L}) \]
\[ = -a^\nu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\nu\phi - \delta^\mu{}_\nu\mathcal{L}\right] \]

Since \(a^\nu\) is an arbitrary constant vector, we obtain an independent conserved current for each \(\nu\). Setting \(j^\mu = -a^\nu T^\mu{}_\nu\):

\[ \boxed{T^{\mu}{}_\nu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\nu\phi - \delta^\mu{}_\nu\,\mathcal{L}} \]

The conservation law is \(\partial_\mu T^{\mu}{}_\nu = 0\) (for each \(\nu\)).

(c) Computation of \(T^{00}\) for the Klein-Gordon field

For \(\mathcal{L} = \frac{1}{2}\partial_\alpha\phi\,\partial^\alpha\phi - \frac{1}{2}m^2\phi^2\):

\[ T^{0}{}_{0} = \frac{\partial\mathcal{L}}{\partial(\partial_0\phi)}\,\partial_0\phi - \delta^0{}_{0}\,\mathcal{L} \]

Since \(\frac{\partial\mathcal{L}}{\partial(\partial_0\phi)} = \partial^0\phi = \dot{\phi}\):

\[ T^{0}{}_{0} = \dot{\phi}\cdot\dot{\phi} - \mathcal{L} = \dot{\phi}^2 - \left[\frac{1}{2}\dot{\phi}^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2\right] \]
\[ = \dot{\phi}^2 - \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]
\[ T^{00} = \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]

Comparing with the Hamiltonian density obtained in S1(b) (using \(\pi = \dot{\phi}\)):

\[ \mathcal{H} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 = \frac{1}{2}\dot{\phi}^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 \]
\[ \boxed{T^{00} = \mathcal{H}} \]

Consistency check: \(T^{00}\) is the energy density and should coincide with the Hamiltonian density. Moreover, \(T^{00} \geq 0\) (each term is positive definite), confirming that the energy is bounded from below. Furthermore, computing \(T^{0i}\) gives \(T^{0i} = \dot{\phi}\,\partial_i\phi\), which corresponds to the momentum density.

M-4. Euler-Lagrange Equations for the Dirac Field

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(a) Euler-Lagrange Equation with Respect to \(\bar{\psi}\)

\[ \mathcal{L}_{\mathrm{Dirac}} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi = i\bar{\psi}\gamma^\mu\partial_\mu\psi - m\bar{\psi}\psi \]

We treat \(\bar{\psi}\) as an independent variable. There is no term containing \(\partial_\mu\bar{\psi}\) in \(\mathcal{L}\) (the \(\partial_\mu\) acts only on \(\psi\)). Therefore:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})} = 0 \]

The Euler-Lagrange equation is:

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})}\right) - \frac{\partial\mathcal{L}}{\partial\bar{\psi}} = 0 \implies -\frac{\partial\mathcal{L}}{\partial\bar{\psi}} = 0 \]

Computing the partial derivative with respect to \(\bar{\psi}\):

\[ \frac{\partial\mathcal{L}}{\partial\bar{\psi}} = i\gamma^\mu\partial_\mu\psi - m\psi = (i\gamma^\mu\partial_\mu - m)\psi \]

Therefore:

\[ \boxed{(i\gamma^\mu\partial_\mu - m)\psi = 0} \]

This is the Dirac equation.

(b) Euler-Lagrange Equation with Respect to \(\psi\)

We treat \(\psi\) as an independent variable.

Partial derivative with respect to \(\psi\):

\[ \frac{\partial\mathcal{L}}{\partial\psi} = -m\bar{\psi} \]

(The term \(i\bar{\psi}\gamma^\mu\partial_\mu\psi\) contains \(\partial_\mu\psi\), so it does not contribute to the partial derivative with respect to \(\psi\) itself.)

Partial derivative with respect to \(\partial_\mu\psi\):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)} = i\bar{\psi}\gamma^\mu \]

Substituting into the Euler-Lagrange equation:

\[ \partial_\mu\left(i\bar{\psi}\gamma^\mu\right) - (-m\bar{\psi}) = 0 \]
\[ i(\partial_\mu\bar{\psi})\gamma^\mu + m\bar{\psi} = 0 \]

Rewriting using the left-acting derivative notation \(\overleftarrow{\partial}_\mu\):

\[ \boxed{\bar{\psi}(i\overleftarrow{\partial}_\mu\gamma^\mu + m) = 0} \]

This is the adjoint Dirac equation.

Verification: We take the Dirac conjugate of the Dirac equation \((i\gamma^\mu\partial_\mu - m)\psi = 0\). Taking the Hermitian conjugate and multiplying by \(\gamma^0\) from the right:

\[ [(i\gamma^\mu\partial_\mu - m)\psi]^\dagger\gamma^0 = 0 \]
\[ \psi^\dagger(-i\gamma^{\mu\dagger}\overleftarrow{\partial}_\mu - m)\gamma^0 = 0 \]

Using the relation \(\gamma^0\gamma^{\mu\dagger}\gamma^0 = \gamma^\mu\) (the Hermiticity property of gamma matrices), we obtain \(\bar{\psi}(i\overleftarrow{\partial}_\mu\gamma^\mu + m) = 0\), which is consistent with the result of (b).

Advanced

A-1. Generalization of Noether's Theorem

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(a) Invariance of the Action

The variation of the action is:

\[ \delta S = \int d^4x\,\delta\mathcal{L} = \int d^4x\,\partial_\mu K^\mu \]

Applying the 4-dimensional Gauss's theorem (divergence theorem):

\[ \delta S = \int d^4x\,\partial_\mu K^\mu = \oint_{\partial\Omega} K^\mu\,dS_\mu \]

where \(\partial\Omega\) is the boundary surface of the 4-dimensional integration region. Under the boundary condition that the field variation \(\delta\phi\) vanishes at the boundary (or that \(K^\mu\) decays sufficiently rapidly at the boundary):

\[ \boxed{\delta S = 0} \]

Therefore, the equations of motion (the Euler-Lagrange equations derived from \(\delta S = 0\)) remain unchanged.

(b) Proof of Conservation of the Modified Noether Current

Computing \(\delta\mathcal{L}\) through the field variation:

\[ \delta\mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi}\,\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\mu(\delta\phi) \]

Rewriting the second term via integration by parts:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\partial_\mu(\delta\phi) = \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\delta\phi \]

Substituting:

\[ \delta\mathcal{L} = \left[\frac{\partial\mathcal{L}}{\partial\phi} - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\right]\delta\phi + \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) \]

When the equations of motion hold, the term in square brackets vanishes:

\[ \delta\mathcal{L} = \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) \]

On the other hand, by assumption \(\delta\mathcal{L} = \partial_\mu K^\mu\). Equating both sides:

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi\right) = \partial_\mu K^\mu \]
\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu\right) = 0 \]
\[ \boxed{\partial_\mu j^\mu = 0, \qquad j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu} \]

(c) Application to Lorentz Boost Transformations

Under an infinitesimal Lorentz boost in the \(x\) direction (rapidity \(\delta\omega\)):

\[ \delta x^0 = -\delta\omega\,x^1, \qquad \delta x^1 = -\delta\omega\,x^0, \qquad \delta x^2 = \delta x^3 = 0 \]

From the scalar field transformation rule \(\delta\phi = -\delta x^\nu\partial_\nu\phi\):

\[ \delta\phi = -(-\delta\omega\,x^1)\partial_0\phi - (-\delta\omega\,x^0)\partial_1\phi = \delta\omega(x^1\partial_0\phi + x^0\partial_1\phi) \]

That is, \(\delta\phi = \delta\omega(x\,\dot{\phi} + t\,\partial_x\phi)\). (This is consistent with the sign convention in the problem statement \(\delta\phi = -\delta\omega(t\,\partial_x\phi + x\,\partial_t\phi)\), where \(x^0 = t\), \(x^1 = x\).)

Determination of \(K^\mu\):

Since \(\mathcal{L}\) is also a scalar, under the same transformation:

\[ \delta\mathcal{L} = -\delta x^\nu\partial_\nu\mathcal{L} = \delta\omega(x^1\partial_0\mathcal{L} + x^0\partial_1\mathcal{L}) \]

Writing this in total derivative form:

\[ \delta\mathcal{L} = \delta\omega\left[\partial_0(x^1\mathcal{L}) + \partial_1(x^0\mathcal{L})\right] - \delta\omega\left[(\partial_0 x^1)\mathcal{L} + (\partial_1 x^0)\mathcal{L}\right] \]

Since \(\partial_0 x^1 = 0\) and \(\partial_1 x^0 = 0\):

\[ \delta\mathcal{L} = \delta\omega\,\partial_\mu\left[\delta^\mu{}_0\,x^1\mathcal{L} + \delta^\mu{}_1\,x^0\mathcal{L}\right] = \partial_\mu K^\mu \]

where \(K^\mu = \delta\omega(\delta^\mu{}_0\,x^1\mathcal{L} + \delta^\mu{}_1\,x^0\mathcal{L})\), i.e., \(K^0 = \delta\omega\,x^1\mathcal{L}\), \(K^1 = \delta\omega\,x^0\mathcal{L}\), \(K^2 = K^3 = 0\).

Construction of the conserved current:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\,\delta\phi - K^\mu \]

Using \(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi\):

\[ j^\mu = \partial^\mu\phi\cdot\delta\omega(x^1\partial_0\phi + x^0\partial_1\phi) - \delta\omega(\delta^\mu{}_0\,x^1\mathcal{L} + \delta^\mu{}_1\,x^0\mathcal{L}) \]

Factoring out \(\delta\omega\):

\[ j^\mu = \partial^\mu\phi(x^1\partial_0\phi + x^0\partial_1\phi) - \delta^\mu{}_0\,x^1\mathcal{L} - \delta^\mu{}_1\,x^0\mathcal{L} \]

Rewriting in terms of the energy-momentum tensor \(T^{\mu}{}_\nu = \partial^\mu\phi\,\partial_\nu\phi - \delta^\mu{}_\nu\mathcal{L}\):

\[ j^\mu = x^1(\partial^\mu\phi\,\partial_0\phi - \delta^\mu{}_0\mathcal{L}) + x^0(\partial^\mu\phi\,\partial_1\phi - \delta^\mu{}_1\mathcal{L}) \]
\[ = x^1 T^{\mu 0} + x^0 T^{\mu 1} \]

Here we used \(T^{\mu\nu} = \eta^{\nu\alpha}T^\mu{}_\alpha\). Noting that \(T^{\mu 0} = T^\mu{}_0\) (since \(\eta^{00} = 1\)) and \(T^{\mu 1} = -T^\mu{}_1\) (since \(\eta^{11} = -1\)):

\[ j^\mu = x^1 T^{\mu 0} - x^0 T^{\mu 1} \]

Alternatively, as the conserved current corresponding to the generator of the Lorentz transformation:

\[ \boxed{M^{\mu 01} = x^0 T^{\mu 1} - x^1 T^{\mu 0}} \]

(The sign convention follows \(M^{\mu\rho\sigma} = x^\rho T^{\mu\sigma} - x^\sigma T^{\mu\rho}\).)

Conserved charge:

\[ M^{01} = \int d^3x\,M^{001} = \int d^3x\,(x^0 T^{01} - x^1 T^{00}) = t\,P^1 - \int d^3x\,x\,\mathcal{H} \]

where \(P^1 = \int d^3x\,T^{01}\) is the total momentum in the \(x\) direction.

Physical meaning: This conserved charge is the boost generator and contains information about the "center of energy" of the system. The conservation of \(M^{01} = tP^1 - \int d^3x\,x\,\mathcal{H}\) means:

\[ \frac{d}{dt}\left(tP^1 - \int d^3x\,x\,\mathcal{H}\right) = 0 \]
\[ P^1 + t\frac{dP^1}{dt} - \frac{d}{dt}\int d^3x\,x\,\mathcal{H} = 0 \]

Using the conservation of \(P^1\) (\(\frac{dP^1}{dt} = 0\)), we obtain \(\frac{d}{dt}\int d^3x\,x\,\mathcal{H} = P^1\). This means that "the center of energy moves at constant velocity," which is the field theory version of the fact that in particle mechanics, \(X_{\mathrm{cm}} = \frac{\sum m_i x_i}{\sum m_i}\) undergoes uniform rectilinear motion.

Verification: We directly confirm \(\partial_\mu M^{\mu\rho\sigma} = 0\). \(\partial_\mu M^{\mu\rho\sigma} = \partial_\mu(x^\rho T^{\mu\sigma} - x^\sigma T^{\mu\rho}) = \delta^\rho_\mu T^{\mu\sigma} + x^\rho\partial_\mu T^{\mu\sigma} - \delta^\sigma_\mu T^{\mu\rho} - x^\sigma\partial_\mu T^{\mu\rho} = T^{\rho\sigma} - T^{\sigma\rho}\) (using \(\partial_\mu T^{\mu\nu} = 0\)). This is conserved if the canonical energy-momentum tensor is symmetric, \(T^{\rho\sigma} = T^{\sigma\rho}\). For the Klein-Gordon field, \(T^{\mu\nu} = \partial^\mu\phi\,\partial^\nu\phi - \eta^{\mu\nu}\mathcal{L}\) is indeed symmetric, which is consistent.

A-2. Lagrangian of the Maxwell Field and Gauge Invariance

→ Back to problem

(a) Derivation of the Euler-Lagrange Equations

\(\mathcal{L}_{\mathrm{Maxwell}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\) with \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\).

Step 1: Computing \(\frac{\partial\mathcal{L}}{\partial A_\nu}\)

Since \(\mathcal{L}\) does not contain \(A_\nu\) itself (without derivatives):

\[ \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 \]

Step 2: Computing \(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\)

Expanding \(F_{\mu\nu}F^{\mu\nu}\):

\[ F_{\alpha\beta}F^{\alpha\beta} = F_{\alpha\beta}\,\eta^{\alpha\gamma}\eta^{\beta\delta}F_{\gamma\delta} = (\partial_\alpha A_\beta - \partial_\beta A_\alpha)\eta^{\alpha\gamma}\eta^{\beta\delta}(\partial_\gamma A_\delta - \partial_\delta A_\gamma) \]

Computing the partial derivative with respect to \(\partial_\mu A_\nu\). Since \(F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha\):

\[ \frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = \delta^\mu_\alpha\delta^\nu_\beta - \delta^\mu_\beta\delta^\nu_\alpha \]

Using the product rule:

\[ \frac{\partial}{\partial(\partial_\mu A_\nu)}(F_{\alpha\beta}F^{\alpha\beta}) = \frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)}F^{\alpha\beta} + F_{\alpha\beta}\frac{\partial F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)} \]

Since \(F^{\alpha\beta} = \eta^{\alpha\gamma}\eta^{\beta\delta}F_{\gamma\delta}\), the second term has the same structure. By symmetry, both terms are equal:

\[ \frac{\partial}{\partial(\partial_\mu A_\nu)}(F_{\alpha\beta}F^{\alpha\beta}) = 2(\delta^\mu_\alpha\delta^\nu_\beta - \delta^\mu_\beta\delta^\nu_\alpha)F^{\alpha\beta} = 2(F^{\mu\nu} - F^{\nu\mu}) = 4F^{\mu\nu} \]

where we used \(F^{\mu\nu} = -F^{\nu\mu}\) (antisymmetry).

Therefore:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\frac{1}{4}\cdot 4F^{\mu\nu} = -F^{\mu\nu} \]

Step 3: The Euler-Lagrange Equations

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\right) - \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 \]
\[ \partial_\mu(-F^{\mu\nu}) - 0 = 0 \]
\[ \boxed{\partial_\mu F^{\mu\nu} = 0} \]

These are the Maxwell equations in vacuum (source-free). Written in components, \(\nu = 0\) corresponds to \(\nabla\cdot\mathbf{E} = 0\) (Gauss's law), and \(\nu = i\) corresponds to \(-\frac{\partial\mathbf{E}}{\partial t} + \nabla\times\mathbf{B} = 0\) (Ampère's law).

(b) Proof of Gauge Invariance

Computing the change in the field strength tensor under the gauge transformation \(A_\mu \to A'_\mu = A_\mu + \partial_\mu\Lambda\):

\[ F'_{\mu\nu} = \partial_\mu A'_\nu - \partial_\nu A'_\mu = \partial_\mu(A_\nu + \partial_\nu\Lambda) - \partial_\nu(A_\mu + \partial_\mu\Lambda) \]
\[ = \partial_\mu A_\nu + \partial_\mu\partial_\nu\Lambda - \partial_\nu A_\mu - \partial_\nu\partial_\mu\Lambda \]

By the commutativity of partial derivatives \(\partial_\mu\partial_\nu\Lambda = \partial_\nu\partial_\mu\Lambda\):

\[ F'_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu = F_{\mu\nu} \]
\[ \boxed{F'_{\mu\nu} = F_{\mu\nu}} \]

Since \(F_{\mu\nu}\) is gauge invariant, \(\mathcal{L}_{\mathrm{Maxwell}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\) is also gauge invariant.

(c) Relationship Between Noether's Theorem and Gauge Transformations

Analysis of the Issue:

Noether's first theorem applies to global continuous symmetries, i.e., transformations whose parameters are constants that do not depend on spacetime. However, the parameter \(\Lambda(x)\) of the gauge transformation \(A_\mu \to A_\mu + \partial_\mu\Lambda(x)\) is an arbitrary function of spacetime (a local parameter).

If we attempt to directly apply Noether's theorem:

  1. Global gauge transformation (\(\Lambda = \text{const}\)): In this case \(\delta A_\mu = \partial_\mu\Lambda = 0\), so the field does not change. Therefore, only the trivial identity \(0 = 0\) is obtained, and no non-trivial conserved current can be constructed.

  2. Local gauge transformation (\(\Lambda(x)\) is an arbitrary function): Since the parameter has infinitely many degrees of freedom, it cannot be handled within the framework of the ordinary Noether theorem.

Noether's Second Theorem:

When the transformation parameters are arbitrary functions of spacetime (local symmetry), Noether's second theorem applies. This theorem states:

If the action is invariant under transformations involving an arbitrary function \(\Lambda(x)\) and its finite-order derivatives, then there exist identities (analogous to Bianchi identities) among the Euler-Lagrange equations.

In the case of Maxwell theory, this identity is:

\[ \partial_\nu(\partial_\mu F^{\mu\nu}) \equiv 0 \]

which holds identically due to the antisymmetry of \(F^{\mu\nu}\), without using the equations of motion.

Physical Meaning:

  • Gauge symmetry is not a physical symmetry (one that generates new conserved quantities), but rather reflects a redundancy in the description of the theory.
  • Of the 4 components of \(A_\mu\), only 2 are physical degrees of freedom (the two polarization states of the photon), and the remaining 2 correspond to gauge redundancy.
  • While Noether's first theorem gives "symmetry → conserved quantity," Noether's second theorem gives "local symmetry → identities among equations of motion (constraints)."
\[ \boxed{\text{Gauge symmetry (local symmetry)} \xrightarrow{\text{Noether's 2nd theorem}} \text{Identities among equations of motion + reduction of physical degrees of freedom}} \]

Verification: Counting the physical degrees of freedom in Maxwell theory. \(A_\mu\) has 4 components, but due to gauge freedom (one arbitrary function \(\Lambda(x)\)), 1 component is redundant. Furthermore, \(\partial_\mu F^{\mu 0} = 0\) provides a constraint (a condition on initial values), removing another component. The result is \(4 - 1 - 1 = 2\) degrees of freedom, corresponding to the two physical polarizations (transverse waves) of the photon.